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Sample Solutions
AP Physics B

Hecht:  Physics with Algebra/Trig

Chapter

Problems
12 pg 437-8 #23, 25, 26, 37

Problems uparrowfont2.gif (884 bytes)

 

#23 f = 0.158 hz

Use T = 2p [ L / g] to calculate the period of the pendulum and then remember that f = 1/T.

T = 2p [ L / g]
T = 2p
[ 10 / 9.8]
T = 2p
1.02
T = 6.35 seconds
f = 0.158 hz

#25 L = 24.8 meters

Use T = 2p [ L / g] to calculate the length of the pendulum.

T = 2p [ L / g]
10.0 = 2p
[ L / 9.8]
102 = 4p2 [ L / 9.8]
100(9.8)/(4p2) = L
L = 24.8 meters

#26 a = 0, b = 0.5 and c = - 0.5

An estimate might be that the period of a simple pendulum might be proportional to its mass m, length L, and the acceleration due to gravity g. Consequently, we will start with the relationship that

T = CmaLbgc

where C is a dimensionless constant.

T =   CmaLbgc
sec1   kga mb [m/sec2]c
kg0m0sec1   kga mb [mc/sec2c]
kg0m0sec1   kga mb+c sec-2c
     
kg:   0 = a
m:   0 = b + c b = -c b = -(-0.5) = 0.5
sec:   1 = -2c c = -0.5

T = C m0 L0.5 g -0.5
T = C[ L / g]

#37 T2 = 1.22 T1

T1 = 2p [ L / g]
T2 = 2p
[ 1.5L / g]
T2 =
1.5  (2p [L / g] )
T2 =
1.22  (2p [L / g] )
T2 = 1.22 T1

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